The coordinates of the point on the line $x+y+3=0$,whose distance from the line $x+2y+2=0$ is $\sqrt{5}$ units,are

Identify the point on the line $2x + 3y + 7 = 0$,which is at a distance of $3$ units from $(1, -3)$.

If $2p$ is the length of the perpendicular from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$,then $a^2, 8p^2, b^2$ are in

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

The number of lines that can be drawn through the point $(4, -5)$ at a distance of $10$ units from the point $(1, 3)$ is

If the line $2x - y + 3 = 0$ is at a distance of $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$ from the lines $4x - 2y + \alpha = 0$ and $6x - 3y + \beta = 0$ respectively,then the sum of all possible values of $\alpha$ and $\beta$ is: